## Perimeter of triangle

Perimeter is the length of the boundary around any closed figure.

The boundary of a triangle consists of its three sides. Therefore, sum of length of three sides of triangle is
called its perimeter.

### What are the units of perimeter?

Units of perimeter measurement of a triangle are taken the same as units of length of sides of the triangle.

The few examples of units of length on the side of a triangle are meter (m), centimeter (cm) etc..
So, the units of perimeter are also taken as meter (m), centimeter (cm) etc..

### How to calculate perimeter of a triangle?

In this ΔABC, we have

length of AB = a

length of BC = b

and length of CA = c

So, perimeter of ΔABC is sum of length of sides AB, BC and CA
**i.e. perimeter of ΔABC** = a + b + c

## Perimeter of an equilateral triangle

In an equilateral triangle, where all sides are of equal length.

i.e. AB = a, BC = a and CA = a

So, **perimeter of equilateral ΔABC** = a + a + a = 3a

## Perimeter of isosceles triangle

In isosceles triangle, where any two sides are of equal length.

i.e. AB = a, BC = a and CA = b

So, **perimeter of isosceles ΔABC** = a + a + b = 2a + b

**Find perimeter of ΔABC with length of its sides as given below:**

AB=2cm, BC=4cm, CA=6cm

So, perimeter of ΔABC = AB + BC + CA

= 2 + 4 + 6

= 12cm

## Area of triangle

Area is the total amount of space occupied by any closed figure.

### What are the units of area?

Area is measured in square units.

The few examples of units of area of triangle are meter^{2} or m^{2} read as meter square,
centimeter^{2} or cm^{2} read as centimeter square etc.

## How to calculate area of a triangle?

Area of any triangle = $\frac{1}{2}$ × base × height

Let’s understand it from following ΔABC

where, BC is base, length of BC = b

and AO is height, length of AO = h

Therefore, area of ΔABC = $\frac{1}{2}$ × b × h

## Area of equilateral triangle

In above equilateral ΔABC, where all sides of triangle are equal in length, its area is calculated
as:

area of ΔABC = $\frac{$ \sqrt{3}$}{4}$
a^{2}
**Let’s see how it is calculated?**

Here, in right angle ΔAOC

a^{2} = h^{2} + ($\frac{\mathrm{a}}{2}$)^{2}
by
Pythagoras theorem

a^{2} = h^{2} +
$\frac{\mathrm{$ {a}^{2}$}}{4}$

a^{2} –
$\frac{\mathrm{$ {a}^{2}$}}{4}$
= h^{2}

$\frac{\mathrm{$ {\mathrm{4a}}^{2}\u2013{a}^{2}$}}{4}$
= h^{2}

$\frac{\mathrm{$ {\mathrm{3a}}^{2}$}}{4}$ = h^{2}

$\frac{$ \sqrt{3}$}{4}$ a = h

Area of Δ =
$\frac{1}{2}$
× a ×
$\frac{$ \sqrt{3}$}{2}$ a

Area of Δ = $\frac{$ \sqrt{3}$}{4}$
a^{2}

## Area of isosceles triangle

area of isosceles ΔABC =
$\frac{\text{b}\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{4}$
**Let’s see how it is calculated?**

Here, in right angle ΔAOC

(a)^{2} = (h)^{2} + ($\frac{\mathrm{b}}{2}$)^{2}
by
Pythagoras theorem

a^{2} = h^{2} +
$\frac{\mathrm{$ {b}^{2}$}}{4}$

a^{2} – $\frac{\mathrm{$ {b}^{2}$}}{4}$ = h^{2}

$\frac{\mathrm{$ {\mathrm{4a}}^{2}\u2013{b}^{2}$}}{4}$
= h^{2}

$\sqrt{\frac{{\mathrm{4a}}^{2}\u2013{b}^{2}}{4}}$ = h

$\frac{\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{2}$ = h

we know, area of Δ = $\frac{1}{2}$ × b × h

∴ area of ΔABC = $\frac{1}{2}$ × b ×
$\frac{\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{2}$

or area of ΔABC =
$\frac{\text{b}\sqrt{{\mathrm{4a}}^{2}\u2013{b}^{2}}}{4}$